M3 June 2005 Q2

2. A closed container \(C\) consists of a thin uniform hollow hemispherical bowl of radius \(a\), together with a lid. The lid is a thin uniform circular disc, also of radius \(a\). The centre \(O\) of the disc coincides with the centre of the hemispherical bowl. The bowl and its lid are made of the same material.

(a) Show that the centre of mass of \(C\) is at a distance \(\tfrac{1}{3}a\) from \(O\). (4)

The container \(C\) has mass \(M\). A particle of mass \(\tfrac{1}{2}M\) is attached to the container at a point \(P\) on the circumference of the lid. The container is then placed with a point of its curved surface in contact with a horizontal plane. The container rests in equilibrium with \(P\), \(O\) and the point of contact in the same vertical plane.

(b) Find, to the nearest degree, the angle made by the line \(PO\) with the horizontal. (5)

Mark scheme (a)

Scheme	Marks
\(\begin{array}{lccc} & \text{Bowl} & \text{Lid} & C \\ \text{Mass ratio} & 2 & 1 & 3 \\ \bar{y} & \tfrac{1}{2}a & 0 & \bar{y} \end{array}\) anything in ratio \(2:1:3\)	B1 B1
M(\(O\)) \(2 \times \tfrac{1}{2}a = 3\bar{y}\)	M1
\(\bar{y} = \tfrac{1}{3}a\) * cso	A1
	(4)

Force diagram: container resting at A with R up, Mg down through the centre of mass, one half Mg down at P, angle theta between OP and the horizontal — Mark scheme (b)

Diagram: O, G-dash on the vertical through O, distances one third a and two ninths a, angle theta — Mark scheme (b)

Scheme	Marks
M(\(A\)) \(Mg \times \tfrac{1}{3}a\sin\theta = \tfrac{1}{2}Mg \times a\cos\theta\)	M1 A1=A1
\(\tan\theta = \tfrac{3}{2}\)	M1
\(\theta \approx 56^\circ\) cao	A1
	(5)
	(9 marks)

\(\begin{array}{lccc} & C & P & C \text{ and } P \\ \text{Mass ratios} & 2 & 1 & 3 \\ \bar{y} & \tfrac{1}{3}a & 0 & \bar{y} \\ \bar{x} & 0 & a & \bar{x} \end{array}\)
Finding both coordinates of \(G^{\prime}\)	M1
\(\tfrac{2}{3}a = 3\bar{y} \Rightarrow \bar{y} = \tfrac{2}{9}a\)	A1
\(a = 3\bar{x} \Rightarrow \bar{x} = \tfrac{1}{3}a\)	A1
\(\tan\theta = \dfrac{\frac{1}{3}a}{\frac{2}{9}a} = \dfrac{3}{2}\)	M1
\(\theta \approx 56^\circ\) cao	A1

\(GG^{\prime} : G^{\prime}P = \tfrac{1}{2}M : M = 1 : 2\)
\(OG = \tfrac{1}{3}a, \quad OP = a\)
By similar triangles \(ON = \tfrac{1}{3}OP = \tfrac{1}{3}a\)	M1 A1
\(NG^{\prime} = \tfrac{2}{3}OG = \tfrac{2}{9}a\)	A1
\(\tan\theta = \dfrac{ON}{NG^{\prime}} = \dfrac{\frac{1}{3}a}{\frac{2}{9}a} = \dfrac{3}{2}\)	M1
\(\theta \approx 56^\circ\) cao	A1

M3 June 2005 Q2

M3 June 2005 Q2

Notes

First Alternative

Second Alternative